Maximum possible connected component by removing hyperplanes form $\mathbb R^3$

Question

Let $A_1,A_2,A_3,A_4$ be 4 hyperplanes in $\mathbb R^3$ then How many maximum possible connected component are present in $\mathbb R^3$ after removing these 4 hyperplane

I can visuallise that if i remove 3 hyperplanes then maximum 8 connected component present.

I donot able to visualise for 4 hyperplanes .

Any help will be appreciated

Answer 1

In some sense, this is akin to removing three lines from $\Bbb R^2$, where you can get at most seven planar regions.

Consider a tetrahedron $T$ in $\Bbb R^3$. By removing the four planes forming the boundary of $T$ one gets fifteen regions remaining.

Let's consider what happens when we remove the fourth plane $P$ in the general case. Before then there are at most eight regions. We claim we can gain at most seven regions at this stage. The earlier planes meet $P$ in at most three lines. These lines divide the plane $P$ into at most seven plane regions. But these plane regions correspond to regions in $\Bbb R^3$ that are split in two by the fourth hyperplane: the number of regions gained by the fourth plane is the number of plane regions the lines divide $P$ into, and is at most seven.